Let $p$ be a prime number $(p \neq 2$ and $p \neq5)$, and let $A$ be some given number. Suppose that $p$ divides the number $A^2 - 5$. Show that $p$ must be congruent to either 1 or 4 modulo 5.
A little confused about this number theory question. Any Help? I would love to see a solution to this problem. Thanks.
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Let $p$ be prime, $p\ne2,5$. We have $p\mid A^2-5$ if and only if $5$ is a quadratic residue modulo $p$, that is, $$\Bigl(\frac5p\Bigr)=1\ .$$ Since $5\equiv1\pmod4$, quadratic reciprocity shows that this is equivalent to $$\Bigl(\frac p5\Bigr)=1\ ,$$ and by various methods (easiest in this case: just trial and error), this is true if and only if $p\equiv1$ or $4$ modulo $5$.
Since $p|A^2-5$, we can say $A^2\equiv 5\pmod{p}$. That means $5$ is a quadratic residue, modulo $p$, or in terms of the Legendre symbol, that $\left(\frac{5}{p}\right)=1$. Since $5$ is congruent to $1$ modulo $4$, we know from quadratic reciprocity that $\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)$. From here, it's just a matter of verifying that quadratic residues modulo $5$ are all congruent to $1$ or $4$.